= Solution
In a neutral <Moran model>, a population has $N$ labelled individuals; reproduction copies one individual's type to a uniformly chosen other individual, keeping population size fixed. One convenient time convention puts independent rate-$1/2$ Poisson arrows on each ordered pair of distinct individuals. Backwards in time, two sampled ancestral lineages merge whenever one copies the other, at total rate $1$ per unordered pair. In the convention where each individual reproduces at rate $1$, this pair rate is $2/(N-1)$ if the parent cannot replace itself, and time must be rescaled accordingly.
<Kingman's coalescent> on $n$ labels is the <continuous-time Markov chain> on partitions of $\{1,\ldots,n\}$, initially all singleton blocks, in which each pair of distinct blocks merges at rate $1$. With $j$ blocks, the waiting time is exponential of rate $\binom j2$, and the merging pair is uniform. <Kingman's infinite coalescent> is the consistent partition process on $\mathbb N$ whose restriction to each finite label <set> has this law; consistency and the extension theorem construct it. These time conventions are part of the definition.
Let $B_t^{(n)}$ be the number of blocks in the first $n$ labels. The expected time to reach at most $m$ blocks is
$$
\mathbb E T_m^{(n)}=\sum_{j=m+1}^n\binom j2^{-1}=2\left(\frac1m-\frac1n\right)\leq\frac2m.
$$
<Markov inequality> gives $\mathbb P(B_t^{(n)}>m)\leq2/(mt)$. As $n$ increases these counts increase to the total number of infinite-coalescent blocks, so the same bound holds for that number. Letting $m\to\infty$ shows it is finite <almost surely> for each $t>0$. Apply this simultaneously to positive rational times; monotonicity of the block count then covers every positive real time. Hence \b[<Kingman's infinite coalescent> comes down from infinity <almost surely>].
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